Updated problem 1 algorithm

Problem1.lua

@@ -1,7 +1,7 @@
 --ProjectEuler/lua/Problem1.lua
 --Matthew Ellison
 -- Created: 02-01-19
---Modified: 06-19-20
+--Modified: 10-26-20
 --What is the sum of all the multiples of 3 or 5 that are less than 1000
 --All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
 --[[
@@ -25,19 +25,23 @@
 require "Stopwatch"
 
 
+local TOP_NUMBER = 999;	--This is the largest number that you are going to check
+local sumOfMultiples = 0;
+
+
 local timer = Stopwatch:create();
 timer:start();
 
-local TOP_NUMBER = 999;	--This is the largest number that you are going to check
+--Gets the sum of the progression of the multiple
-local sumOfMultiples = 0;
+local function sumOfProgression(multiple)
-for num = 1, TOP_NUMBER do
+	local numTerms = TOP_NUMBER // multiple;	--This gets the number of multiples of a particular number that is < MAX_NUMBER
-	if((num % 3) == 0) then
+	--The sum of progression formula is (n / 2)(a + l). n = number of terms, a = multiple, l = last term
-		sumOfMultiples = sumOfMultiples + num;
+	return ((numTerms / 2) * (multiple + (numTerms * multiple)));
-	elseif((num % 5) == 0) then
-		sumOfMultiples = sumOfMultiples + num;
-	end
 end
 
+--Get the sum of the progressions of 3 and 5 and remove the sum of progressions of the overlap
+sumOfMultiples = math.floor(sumOfProgression(3)) + math.floor(sumOfProgression(5)) - math.floor(sumOfProgression(3 * 5));
+
 timer:stop()
 
 --Print the results
@@ -47,5 +51,5 @@ print("It took " .. timer:getMicroseconds() .. " microseconds to run this algori
 
 --[[Results:
 The sum of all numbers < 1000 is 233168
-It took 148 microseconds to run this algorithm
+It took 2.0 microseconds to run this algorithm
 ]]